Momentum with Two Carts on a Low-Friction Track

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Homework Statement


A student runs an experiment with two carts on a low-friction track. As measured in the Earth reference frame, cart 1 (m = 0.48 kg ) moves from left to right at 1.0 m/s as the student walks along next to it at the same velocity. Let the +x direction be to the right.

A. What velocity v⃗ E2,i in the Earth reference frame must cart 2 (m = 0.16 kg ) have before the collision if, in the student's reference frame, cart 2 comes to rest right after the collision and cart 1 travels from right to left at 0.33 m/s?
B. What does the student measure for the momentum of the two-cart system?
C. What does a person standing in the Earth reference frame measure for the momentum of each cart before the collision?

Homework Equations


(m1)(Ve1x,i) + (m2)(Ve2x,i) = (m1)(Ve1x,f) + (m2)(Ve2x,f)

The Attempt at a Solution


A. (0.48 kg)(1 m/s) + (0.16 kg)(Ve2x,i) = (0.48 kg)(-(1/3)m/s) + (0.16 kg)(0 m/s)
I got Ve2x,i = -2.0 m/s which doesn't make sense and it is in fact wrong. Where am I messing up? Also our teacher said to use (1/3) for the velocity of cart 1 after the collision instead of 0.33. I did try both, 0.33 gave me 3.99 which I entered as 4 and it was wrong.
 
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Check your frame of references. Your relationship is a mixture of velocities in the Earth frame and in the students frame. You must pick a frame and ensure all numbers are for that frame only.
 
rpthomps said:
Check your frame of references. Your relationship is a mixture of velocities in the Earth frame and in the students frame. You must pick a frame and ensure all numbers are for that frame only.
How do you manipulate the equation to do so? I'm sorry we've just started reference frames and I don't totally understand how to make sure the equation distinguishes from the two.
 
Well, the person is walking at 1 m/s and they notice that a cart is stationary after a collision. How fast must the cart be moving in the Earth's frame in order for the person to see this?
 
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